# Find shortest distance between lines in 3D

Find shortest distance between lines given by $x-2/3=y-6/4=z+9/-4$ and $x+1/2=y+2/-6=z-3/1$ Is there any shortcut method for this problems?

• Have you tried anything? Please show your efforts – Lelouch Apr 1 '17 at 15:11
• No I didn't try.I don't know how to solve it. – Amey Patil Apr 1 '17 at 15:15
• Well, see they can either be skew or intersecting. first try to solve them. – samjoe Apr 1 '17 at 15:17

So you have two lines defined by the points $\mathbf{r}_1=(2,6,-9)$ and $\mathbf{r}_2=(-1,-2,3)$ and the (non unit) direction vectors $\mathbf{e}_1=(3,4,-4)$ and $\mathbf{e}_2 =(2,-6,1)$.

The coordinates of all the points along the lines are given by

\begin{align} \mathbf{p}_1 & = \mathbf{r}_1 + t_1 \mathbf{e}_1 \\ \mathbf{p}_2 & = \mathbf{r}_2 + t_2 \mathbf{e}_2 \\ \end{align}

where $t_1$ and $t_2$ are two scalar values. To find the closest points along the lines you recognize that the line connecting the closest points has direction vector $$\mathbf{n} = \mathbf{e}_1 \times \mathbf{e}_2 = (-20,-11,-26)$$

If the points along the two lines are projected onto the cross line the distance is found with one fell swoop

$$d = \frac{ \mathbf{n}\cdot \mathbf{p}_1}{\|\mathbf{n}\|} - \frac{ \mathbf{n}\cdot \mathbf{p}_2}{\|\mathbf{n}\|} = \frac{ \mathbf{n} \cdot ( \mathbf{p}_1-\mathbf{p}_2)}{\| \mathbf{n} \|} = \frac{ \mathbf{n} \cdot ( \mathbf{r}_1-\mathbf{r}_2+t_1 \mathbf{e}_1 -t_2 \mathbf{e}_2)}{\| \mathbf{n} \|}$$

But since $\mathbf{n}\cdot \mathbf{e}_1 = \mathbf{n}\cdot \mathbf{e}_2 = 0$, the above is

$$\boxed{ d = \frac{ \mathbf{n} \cdot ( \mathbf{r}_1-\mathbf{r}_2)}{\| \mathbf{n} \|} }$$

In this case $$d = \frac{ (-20,-11,-26) \cdot (3,8,-12) }{3 \sqrt{133}} = 4.74020116673185$$

• . . . nice . . . – janmarqz Nov 30 '18 at 22:53

Here is an approach that uses dot products instead of cross products.

This works in any number of dimensions, not just 3.

The skew lines are $L = a+bt, M=c+ds$.

The distance between two points on $L$ and $M$ is $D =(a+bt-c-ds)^2 =(e+bt-ds)^2$ where $e = a-c$.

For this to be a minimum, taking partials, we want $D_s = D_t = 0$.

$D_s = -2d(e+bt-ds)$ and $D_t = 2b(e+bt-ds)$.

Therefore, with multiplication of vectors being dot product, $0 =d(e+bt-ds) =de+dbt-d^2s$ and $0 =b(e+bt-ds) =be+b^2t-bds)$.

These are two equations in the two unknowns $s$ and $t$:

$\begin{array}\\ de &= d^2s-dbt\\ be &= bds-b^2t\\ \end{array}$

The determinant is $A =-b^2d^2+(bd)^2 =-(b^2d^2-(bd)^2)$. By Cauchy-Schwarz, this is non-zero unless $b$ and $d$ are parallel (which is a good thing).

The solutions (by Cramer's rule) are $s =\dfrac{-(b^2)(de)+(be)(db)}{A}$ and $t =\dfrac{(d^2)(be)-(be)(db)}{A}$.

Putting these into $L = a+bt, M = c+ds, D =(e+bt-ds)^2$ we get the endpoints of the closest line and the distance.

• I think that in the solution for $t$ the second $(be)$ term should actually be $(de)$. – Eskapp Feb 27 '18 at 15:38
• Also it is $D^2 = (e + bt - ds)^2$ One needs to take the square root to get the distance. – Eskapp Feb 27 '18 at 20:28

Hint:

write the equations of the two lines in the form $\vec x=\vec p+t\vec q$: $$r_1) \qquad \begin{pmatrix} x\\y\\z \end{pmatrix}=\begin{pmatrix} 2\\6\\-9 \end{pmatrix}+t\begin{pmatrix} 3\\4\\-4 \end{pmatrix}$$ $$r_2) \qquad \begin{pmatrix} x\\y\\z \end{pmatrix}=\begin{pmatrix} -1\\-2\\3 \end{pmatrix}+t\begin{pmatrix} 2\\-6\\1 \end{pmatrix}$$

than, noted the the two lines are not parallel nor intersecting, use the formula from here.